Find the only x, y, z positive integers such that x^3 = (z-y)^3 + 3yz(z-y) Presentation at: https://sleepy-noyce-8b3d2a.netlify.com/#1

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Find the only $x$, $y$, and $z$ positive integers

$$ x^3 = (z - y)^3 + 3yz(z - y) $$

How you solve math problems

Poke around and understand the underlying nature of the problem
Figure out how to best represent it
How is the best way of solving this problem

Poke around

Trying trivial solutions

One common thing I do with all sorts of math problems is try simple values to see if I can figure out some type of pattern within the problem.

Ex. $0, 1, -1, 5, 10, e, \pi$

Try $1$

$$ x = 1, y = 1, z = 1 $$

which leads to

$$ 1^3 = (1-1)^3 + 3 \times 1 \times 1 \times (1 - 1) \\ 1 = 0^3 + 3 \times 1 \times 1 \times 0 \\ 1 = 0 + 0 \\ 1 = 0 $$

But that's not really true. Setting everything to 1 does not work.

If you look carefully because of the $z -y$ terms if $y = z$ then the right hand side is always zero.

Figure out how to best represent it

Figure out how to best representent it - Expand it

$$ x^3 = z^3 - 3z^2 y + 3z y^2 - y^3 + 3z^2 y - 3 z y^2 $$

Sum and difference cubes formula from highschool if you forgot

Okay I see a bunch of terms that cancel out.

$$ x^3 = z^3 - y^3 $$

One cube is the difference of another set of cubes

$$ x^3 + y^3 = z^3 $$

Now this looks like a cube version of the pythogoras theorem. Hmm... is that relevant. Let's turn to google.

Fermat's last theorem

States that no three positive integers a, b, and c satisfy the equation $a^n + b^n = c^n$ for any integer n greater than 2.

Honestly the rest I can accept you knowing but this isn't common highschool knowledge. Don't worry if you did not know this theorem.

Basically here the problem has $n =3$, so you were given an impossible to solve problem.

Still you have to always poke around the problem and it wasn't immediately obvious that it was one until we dove in. Often in maths you are actually expected to show that a problem isn't solvable. (and even sometimes get awards and massive fame for it)

Okay what if it doesn't have to be positive

Try $0$

$$ x = 0, y = 0, z = 0 $$

which becomes $$ 0^3 + 0^3 = 0^3 \ 0 = 0 $$

which is correct. That's one solution.

Remember when we wondered about $y = z$.

Let's explore that, by subbing $y$ for $z$ $$ x^3 + y^3 = y^3 \ x^3 = y^3 - y^3 \ x^3 = 0 \ x = 0 $$

So if $x = 0$ and $y = z$ then we have infinite solutions (because y and z can be anything so long as they are equal)

A lesson to be learnt

Well honestly I could have just looked at the text and said well this is impossible, but there is always something to learn from every problem and explaining a thought process around this is always a good chance to learn. Math is all about thinking after all

A funny story

So we used Fermat's last theorem, but interestingly it was not proved for 357 years. He had written this super famous result on the side margin saying I have discovered a truly remarkable proof of this theorem which this margin is too small to contain..

It puzzled so many top mathematicians and the one who finally found it was given a massive prize.

Moral of the story: Show your work :)

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